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  "body": "\n## Definition\n\nLet $f$ be a [[smooth function|0BE3QDH73A]] on an open interval that contains $[0, \\infty)$, and let $B_{2k}$ be the [[Bernoulli numbers|0RAS24E1R5]], with $B_2 = \\frac16$.\nAssume that the series $\\sum_{k \\ge 1} \\frac{B_{2k}}{(2k)!} f^{(2k-1)}(0)$ converges.\nThe *Ramanujan sum* of the series $\\sum_{n \\ge 1} f(n)$ is\n$$\\sum_{n \\ge 1}^{\\mathcal{R}} f(n) = -\\frac{f(0)}{2} - \\sum_{k \\ge 1} \\frac{B_{2k}}{(2k)!} f^{(2k-1)}(0).$$\n\n## Remarks\n\nIf $f$ is a polynomial, only finitely many derivatives are nonzero and the series in the definition is a finite sum.\n\nExample: for $f(x) = x$, one has $f(0) = 0$, $f'(0) = 1$ and all higher derivatives vanish.\nSo the Ramanujan sum of $1 + 2 + 3 + \\cdots$ is $-\\frac{B_2}{2} = -\\frac{1}{12}$.\nFor $f(x) = 1$, the Ramanujan sum of $1 + 1 + 1 + \\cdots$ is $-\\frac12$.\n\nNot a regular method: for $f(x) = e^{-x}$ the series $\\sum_{n \\ge 1} e^{-n}$ converges to $\\frac{1}{e - 1}$.\nFrom $\\frac{t}{e^t - 1} = 1 - \\frac{t}{2} + \\sum_{k \\ge 1} \\frac{B_{2k}}{(2k)!} t^{2k}$ at $t = 1$ one gets $\\sum_{k \\ge 1} \\frac{B_{2k}}{(2k)!} = \\frac{1}{e-1} - \\frac12$.\nSince $f^{(2k-1)}(0) = -1$, the Ramanujan sum is $-\\frac12 + \\frac{1}{e-1} - \\frac12 = \\frac{1}{e-1} - 1$.\nIt differs from the ordinary sum by $\\int_0^\\infty e^{-x} \\, dx = 1$.\n\nNot stable: for $g(x) = f(x + 1)$ with $f = 1$, the Ramanujan sum of $\\sum g(n)$ is $-\\frac12$, not $-\\frac12 - f(1) = -\\frac32$.\n",
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    "name": "Ramanujan summation",
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